# Rolle's Theorem Questions and Examples

Rolle's theorem is a special case of the mean value theorem. It is discussed here through examples and questions.

## Rolle's Theorem

Rolle's theorem is the result of the mean value theorem where under the conditions:
$$f(x)$$ be a
continuous functions on the interval $$[a, b]$$ and differentiable on the open interval $$(a, b)$$ , there exists at least one value c of x such that $f '(c) = \dfrac{f(b) - f(a)}{b - a}$ Now if the condition $$f(a) = f(b)$$ is satisfied, then the above simplifies to : $$f'(c) = 0$$.
In other words, under the above three conditions we can always find a tangent to the curve of f that is horizontal (slope = $$f'(c) = 0$$ ).
Rolle's Theorem
If $$f(x)$$ is
1) a continuous function on the interval $$[a, b]$$
2) differentiable on the open interval $$(a, b)$$
3) and $$f(a) = f(b)$$,
then there is at least one value c of x in the interval $$(a, b)$$ such that
$$f'(c) = 0$$

## Example 1

The graph of $$f(x) = - x^2 + 6x - 6$$ for $$1 \leq x \leq 5$$ is shown below. $$f(1) = f(5) = - 1$$ and $$f$$ is continuous on [1 , 5] and differentiable on (1 , 5) hence, according to Rolle's theorem, there exists at least one value of $$x = c$$ such that $$f '(c) = 0$$.
$$f '(x) = - 2 x + 6$$
$$f '(c) = - 2 c + 6 = 0$$
Solve the above equation to obtain
$$c = 3$$
Therefore at $$x = 3$$ there is a tangent to the graph of $$f$$ that has a slope equal to zero (horizontal line) as shown in figure 1 below.

## Example 2

The graph of $$\quad f(x) = \sin(x) + 2$$ for $$\quad 0 \leq x \leq 2\pi$$ is shown below. $$\quad f(0) = f(2\pi) = 2$$ and $$f$$ is continuous on [0 , $$2\pi$$] and differentiable on (0 , $$2\pi$$) hence, according to Rolle's theorem, there exists at least one value ( there may be more than one! ) of $$x = c$$ such that $$f '(c) = 0$$.
$$f '(x) = \cos(x)$$
$$f '(c) = \cos(c) = 0$$
The above equation has two solutions on the interval [0 , $$2\pi$$]
$$c_1 = \dfrac{\pi}{2}$$ and $$c_2 = \dfrac{3\pi}{2}$$.
Therefore both at $$x = \dfrac{\pi}{2}$$ and $$x = \dfrac{3\pi}{2}$$ there are tangents to the graph that have a slope equal to zero (horizontal line) as shown in figure 2 below.

## Example 3

Function $$f$$ in figure 3 does not satisfy Rolle's theorem: although it is continuous and $$f(-1) = f(3)$$, the function is not differentiable at $$x = 1$$ and therefore $$f '(c) = 0$$ with $$c$$ in the interval (-1 , 3) is not guaranteed. In fact it is easy to see that there is no horizontal tangent to the graph of $$f$$ on the interval (-1 , 3).

## Question 1

Which of the functions given below satisfy all three conditions of Rolle's theorem?
a) $$f(x) = \cos(x)$$, for $$x$$ in $$[0 , 2\pi$$] \)
b) $$g(x) = |x - 2|$$, for $$x$$ in $$[0 , 4]$$
c) $$h(x) = 1 / x^2$$, for $$x$$ in $$[-1 , 1]$$
d) $$k(x) = |\sin(x)|$$, for $$x$$ in $$[0 , 2\pi ]$$

### solution to question 1

a)
$$f(0) = 1$$ and $$f(2\pi) = 1$$ therefore $$f(0) = f(2\pi)$$
$$f$$ is continuous on $$[0 , 2\pi ]$$
Function $$f$$ is differentiable in $$(0 , 2\pi )$$
Function $$f$$ satisfies all conditions of Rolle's theorem
b)
function $$g$$ has a V-shaped graph with vertex at $$x = 2$$ and is therefore not differentiable at $$x = 2$$.
Function $$g$$ does not satisfy all conditions of Rolle's theorem
c)
Function $$h$$ is undefined at $$x = 0$$.
Function $$h$$ does not satisfy all conditions of Rolle's theorem.
d)
$$k(x) = |\sin(x)|$$, for $$x$$ in $$[0 , 2\pi]$$
The graph of function $$k$$ is shown below and it shows that function $$k$$ is not differentiable at $$x = \pi$$.
Function $$k$$ does not satisfy all conditions of Rolle's theorem.

## Question 2

Check that function $$f(x) = x^2 - 4x + 3$$ on the interval [1 , 3] satisfies all conditions of Rolle's theorem and then find all values of $$x = c$$ such that $$f '(c) = 0$$.

### solution to question 2

$$f$$ is a polynomial function and is therefore continuous on the interval [1 , 3] and differentiable on the interval (1 , 3). Also $$f(1) = f(3) = 0$$ and therefore function $$f$$ satisfies all three conditions of Rolle's theorem and there is at least one value of $$x = c$$ such that $$f '(c) = 0$$.
$$f '(x) = 2x - 4$$
$$f '(c) = 2c - 4 = 0$$
Solve for $$c$$ to obtain
$$c = 2$$.
The graph below shows $$f$$, $$f'$$ and the tangent at $$x = c = 2$$ is horizontal; and $$f'(2) = 0$$

## Question 3

Check that function $$g(x) = \cos(x)$$ on the interval $$[- \pi /2 , 3\pi/2]$$ satisfies all conditions of Rolle's theorem and then find all values $$x = c$$ such that $$g '(c) = 0$$.

### solution to question 3

Function $$g$$ is a cosine function and is therefore continuous on the interval $$[- \pi/2 , 3\pi/2]$$ and differentiable on the interval $$(- \pi /2 , 3\pi /2)$$ . Also $g(- \pi /2) = g( 3\pi /2) = 0$ and therefore function $$g$$ satisfies all three conditions of Rolle's theorem and there is at least one value of $$x = c$$ such that $$f '(c) = 0$$.
$$g '(x) = - \sin(x)$$
$$g '(c) = - \sin(c) = 0$$
Solve for $$c$$ to obtain
$$c = n \pi$$ , $$n = 0,\pm 1 , \pm 2 , ...$$
Solutions on the interval $$[- \pi/2 , 3\pi /2]$$ are
$$c_1 = 0$$ and $$c_2 = \pi$$;
The graph below shows $$g$$, $$g'$$ and the tangents at $$x = c_1 = 0$$ and $$x = c_2 = \pi$$ are horizontal; and $$g'(0) = 0$$ and $$g'(\pi) = 0$$